rt，如果没写错的话，理论时间复杂度应该是 $O(n\log^2n)$ 的，但是代码没跑过 $10^5$。

#include<iostream>
#define mid ((l+r)/2)
using namespace std;
int n,q;
int a[200005];
struct EDGE{
	int u,v,w,nxt;
}edge[200005];
int head[100005],tot;
void add(int u,int v,int w){
	edge[++tot]={u,v,w,head[u]};
	head[u]=tot;
}
int dep[100005],fa[100005],hson[100005],siz[100005];
void dfs1(int x,int f){
	fa[x]=f;
	dep[x]=dep[f]+1;
	siz[x]=1;
	for(int i=head[x];i;i=edge[i].nxt){
		int v=edge[i].v;
		if(v==f)continue;
		a[v]=edge[i].w;
		dfs1(v,x);
		siz[x]+=siz[v];
		if(siz[v]>siz[hson[x]])hson[x]=v;
	}
}
int top[100005],dfn[100005],idx,rnk[100005];
void dfs2(int x){
	dfn[x]=++idx;
	rnk[idx]=x;
	if(hson[x]){
		top[hson[x]]=top[x];
		dfs2(hson[x]);
	}
	for(int i=head[x];i;i=edge[i].nxt){
		int v=edge[i].v;
		if(v==fa[x]||v==hson[x])continue;
		top[v]=v;
		dfs2(v);
	}
}
struct node{
	int maxn,minn;
}tr[400005];
void pushup(int x){
	tr[x].maxn=max(tr[2*x].maxn,tr[2*x+1].maxn);
	tr[x].minn=min(tr[2*x].minn,tr[2*x+1].minn);
}
void build(int x,int l,int r){
	if(l==r){
		tr[x].maxn=tr[x].minn=a[rnk[l]];
		return ;
	}
	build(2*x,l,mid);
	build(2*x+1,mid+1,r);
	pushup(x);
}
int query(int x,int l,int r,int from,int to,int f){
	if(from<=l&&r<=to){
		if(f==1)return tr[x].maxn;
		return tr[x].minn;
	}
	int ans;
	if(f==1)ans=-1e18;
	else ans=1e18;
	if(from<=mid){
		if(f==1)ans=max(ans,query(2*x,l,mid,from,to,f));
		if(f==2)ans=min(ans,query(2*x,l,mid,from,to,f));
	}
	if(to>mid){
		if(f==1)ans=max(ans,query(2*x+1,mid+1,r,from,to,f));
		if(f==2)ans=min(ans,query(2*x+1,mid+1,r,from,to,f));
	}
	return ans;
}
void Query(int u,int v){
	int ans1=-1e18,ans2=1e18;
	while(top[u]!=top[v]){
		if(dep[top[u]]<dep[top[v]])swap(u,v);
		ans1=max(ans1,query(1,1,n,dfn[top[u]],dfn[u],1));
		ans2=min(ans2,query(1,1,n,dfn[top[u]],dfn[u],2));
		u=fa[top[u]];
	}
	if(dep[u]<dep[v])swap(u,v);
	if(u!=v){
		ans1=max(ans1,query(1,1,n,dfn[v]+1,dfn[u],1));
		ans2=min(ans2,query(1,1,n,dfn[v]+1,dfn[u],2));
	}
	cout<<ans2<<' '<<ans1<<endl;
}
signed main(){
	cin>>n;
	for(int i=1;i<n;i++){
		int u,v,w;
		cin>>u>>v>>w;
		add(u,v,w);
		add(v,u,w);
	}
	dfs1(1,0);
	top[1]=1;
	dfs2(1);
	build(1,1,n);
	cin>>q;
	while(q--){
		int u,v;
		cin>>u>>v;
		Query(u,v);
	}
	return 0;
}